Rotating Space Station Consider out in deep space, a hypothetical space station

Question

Rotating Space Station Consider out in deep space, a hypothetical space station that can be well-approximated as a hoop of radius R = 1000 m and mass M = 850,000 kg. At the edge of the space station there are four equally spaced rocket engines that can be used to make the space station spin, or stop spinning depending on the direction they are pointed. The rocket engines can only fire tangent to the space station. Initially the space station is not rotating, when suddenly the four rockets turn on. Each exerts a force f(t) = -at2 + bt. The rockets exert the force from t = 0 s until the next time the force function equals zero, at which point they instantly shut off. What dimensions and SI-units should a and b have? If a has numerical value 1 and b has numerical value 5, how much angular momentum has the space station gained as a result of the engines firing (assume the rocket exhaust's mass is small, so the station still has mass M)? What is the station's final angular velocity, in radians per second? What is the station's angular acceleration at t = 2 s? What is the total angular displacement of a point on the edge of the station during the period of angular acceleration? We say that angular momentum is a conserved quantity, but the space station appears to gain angular momentum. Reconcile these two observations. If the rocket exhaust consists of particles with mass m = 6 times 10-26 kg and they leave the rocket at a speed of 100,000 m/s. How many of these particles must be in the exhaust for momentum to be conserved?

Explanation / Answer

Part a)

a: N/s^2

b: N/s

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Part b)

moment of inertia:

I = M R^2 = (850000)*(1000*1000) = 8.5e11 Kg.m2

force:

F = -a t^2 + b t

==> F = -t^2 + 5 t

==> torque = T = 4 F R = 4 * (-t^2 + 5 t) * (1000) = -4000 t^2 + 20000 t

==> angular accelertion = alpha = T/I = (-4000 t^2 + 20000 t)/(8.5e11) = -(4.7059e-9) t^2 + (2.3529e-8) t

==> angular speed = w = integral{alpha dt}

==> w = -(4.7059e-9) (t^3/3) + (2.3529e-8) (t^2/2) = -(1.5686e-9) t^3 + (1.17645e-8) t^2

at t=5s the force of the engines is zero and the engines will turn off:

==> angular speed = w = -(1.5686e-9)*(5)^3 + (1.17645e-8)*(5)^2 = 9.80375e-8 rad/s

==> angular momentum = L = I w = (8.5e11) * (9.80375e-8) = 8.3331875e4 = 8.33 x 10^4 Kg.m^2/s

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Part c)

angular speed = w = -(1.5686e-9)*(5)^3 + (1.17645e-8)*(5)^2 = 9.80375e-8 = 9.80 x 10^-8 rad/s

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Part d)

angular accelertion = alpha = T/I = (-4000 t^2 + 20000 t)/(8.5e11) = -(4.7059e-9) t^2 + (2.3529e-8) t

at t=2s:

alpha = -(4.7059e-9)*(2)y2 + (2.3529e-8)*(2)

==> alpha = 2.82 x 10^-8 rad/s^2

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Part e)

theta = integral{w dt} = integral{(-(1.5686e-9) t^3 + (1.17645e-8) t^2) dt}

==> theta = -(1.5686e-9) (t^4/4) + (1.17645e-8) (t^3/3)

==> theta = -(3.9215e-10) t^4 + (3.9215e-9) (t^3)

at t=5s:

==> theta = -(3.9215e-10)*(5^4) + (3.9215e-9)*(5^3)

==> theta = 2.45 x 10^-7 radians

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Part f)

The angular momentum is constant for a closed system but here the station system is not closed because the station loses ass as the exhaust.

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